Ta có : \(\left(x^2-y^2\right)^2+4x^2y^2+x^2-2y^2=0\)

\(\Leftrightarrow\left(x^2+y^2\right)^2-2.\left(x^2+y^2\right)+1=1-3x^2\)

\(\Leftrightarrow\left(x^2+y^2-1\right)^2=1-3x^2\le1\forall x\)

\(\Rightarrow\left(x^2+y^2-1\right)\le1\)

\(\Rightarrow-1\le x^2+y^2-1\le1\)

\(\Rightarrow0\le x^2+y^2\le2\)

\(C=x^2+y^2\) min tại \(x=y=0\)

\(C=x^2+y^2\)max tại \(x=0,y=\sqrt{2}\)

a Tách \(M=2+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\le2+1=3\)
Dấu = xảy ra khi và chỉ khi x=y và x+y=2015 <=>x=y=2015/2
b,:\(N\ge\frac{\left(1+\frac{2015}{x}+1+\frac{2015}{y}\right)^2}{2}=\frac{\left(2+2015\left(\frac{1}{x}+\frac{1}{y}\right)\right)^2}{2}\)
áp dunngj svac =>\(N\ge\frac{\left(2+2015\left(\frac{\left(1+1\right)^2}{x+y}\right)\right)^2}{2}=\frac{\left(2+\frac{2015.4}{2015}\right)^2}{2}=18\)
dấu = xảy ra khi và chỉ khi x=y và x+y=2015 <=>x=y=2015/2

Cảm ơn bn nha :))

@ mình rep lúc 12h mà trả lời lúc 11h :))

\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)

\(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{x^2+2xy+y^2}+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\)

\(\le2+\frac{4.1006^2}{2012^2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)

\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

... 

cảm ơn bạn nhiều

bạn ơi, mik học \(A^2+B^2\ge\left(A+B\right)^2d\text{ấu}"="\) xảy ra <=> \(A.B\ge0\) mà bạn?

pt \(\Leftrightarrow\)\(\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}=-y^2+\frac{49}{4}-10\)

\(\Leftrightarrow\)\(\left(x+y+\frac{7}{2}\right)^2=-y^2+\frac{9}{4}\le\frac{9}{4}\)

\(\Leftrightarrow\)\(\frac{-3}{2}\le x+y+\frac{7}{2}\le\frac{3}{2}\)

\(\Leftrightarrow\)\(-4\le x+y+1\le-1\)

Dấu "=" tự xét nhé 

Ta có: \(2\left(x^2+y^2\right)=1+xy\)

\(\Leftrightarrow x^2+y^2=\frac{1+xy}{2}\)

\(P=7\left(x^4+y^4\right)+4x^2y^2\)

\(=7x^4+7y^4+4x^2y^2\)

\(\Rightarrow P=28x^3+28y^3+16xy\)

\(\Leftrightarrow P=0\Leftrightarrow28x^3+28y^3+16xy=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\y=4\end{cases}}\)

\(\Rightarrow P_{Min}=15\) và \(Max_P=\frac{12}{33}\)

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2\left(x^2+y^2\right)+xy\ge5xy\\2\left(x^2+y^2\right)+xy\ge-3xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\ge5xy\\1\ge-3xy\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{3}\le xy\le\dfrac{1}{5}\)

Ta có:

P=\(2\left(x^2+y^2\right)^2-4x^2y^2+2+\left(x^2+y^2+2xy\right)\)

P= \(\dfrac{2\left(1-xy\right)^2}{4}-4\left(xy\right)^2+2+\left(\dfrac{1-xy}{2}+2xy\right)\)

=\(\dfrac{\left(xy\right)^2-2xy+1}{2}-4\left(xy\right)^2+2+\dfrac{3xy}{2}+\dfrac{1}{2}\)

Đặt t = xy => \(-\dfrac{1}{3}\le t\le\dfrac{1}{5}\)

Ta có : 

P= \(\dfrac{-7t^2}{2}+\dfrac{t}{2}+3=-\dfrac{7}{2}\left(t-\dfrac{1}{14}\right)^2+\dfrac{169}{56}\)

Ta có: \(-\dfrac{1}{3}-\dfrac{1}{14}\le t-\dfrac{1}{14}\le\dfrac{1}{5}-\dfrac{1}{14}\)

<=>\(-\dfrac{17}{42}\le t-\dfrac{1}{14}\le\dfrac{9}{70}\)

=> 0\(\le\left(t-\dfrac{1}{14}\right)^2\le\left(\dfrac{17}{42}\right)^2\)

\(\dfrac{169}{56}\ge P\ge\dfrac{169}{56}-\dfrac{7}{2}\left(\dfrac{17}{42}\right)^2\)

Max P= \(\dfrac{169}{56}\) => t = 1/14 => \(xy=\dfrac{1}{14}\rightarrow x^2+y^2=\dfrac{13}{14}\) => x,y=...

Min P=\(\dfrac{169}{56}-\dfrac{7}{6}\left(\dfrac{17}{42}\right)^2\) <=> \(t=xy=-\dfrac{1}{3}\)

<=> x=-y=\(\dfrac{1}{\sqrt{3}}\)